Uva 216 Getting in Line

最新推荐文章于 2016-07-27 20:23:52 发布

原创最新推荐文章于 2016-07-27 20:23:52 发布 · 310 阅读

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#permutation #network #struct

本文介绍了一种通过暴力枚举所有可能路径来找出连接多个二维坐标点间的最短路径的方法。该方法适用于数据量较小的情况，利用C++实现了一个具体的算法示例，能够输出最优路径及其总长度。

枚举给定点的坐标间距离和的最小值~~数据量小，暴力枚举即可~~

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cctype>
#include<cstdlib>
#include<algorithm>

using namespace std;

typedef struct _Point{
    int xx,yy;
}Point;

Point point[10];
int perm[10],res_perm[10];
double res_dist[10];

int main()
{
    freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int xx,yy;
    int n;
    int cc = 0;
    while(scanf("%d",&n)==1&&n)
    {
        double min = 99999999.999;
        for(int i = 0;i < n;i++)
            scanf("%d %d",&point[i].xx,&point[i].yy);
        for(int i = 0;i < n;i++)    perm[i] = i;
        while(next_permutation(perm,perm+n))
        {
            double dist[n-1];
            for(int i = 1;i < n;i++)
            {
                dist[i-1] = 16+sqrt((point[perm[i]].xx-point[perm[i-1]].xx)*(point[perm[i]].xx-point[perm[i-1]].xx)
                            +(point[perm[i]].yy-point[perm[i-1]].yy)*(point[perm[i]].yy-point[perm[i-1]].yy));
            }
            double cable = 0;
            for(int i = 0;i < n-1;i++)
                cable += dist[i];
            if(cable < min)
            {
                min = cable;
                for(int i = 0;i < n;i++){res_perm[i]=perm[i];}
                for(int i = 0;i < n-1;i++){res_dist[i]=dist[i];}
            }
        }
        printf("**********************************************************\n");
        printf("Network #%d\n",++cc);
        for(int i = 0;i < n-1;i++)
            printf("Cable requirement to connect (%d,%d) to (%d,%d) is %.2lf feet.\n",point[res_perm[i]].xx,point[res_perm[i]].yy,point[res_perm[i+1]].xx,point[res_perm[i+1]].yy,res_dist[i]);
        printf("Number of feet of cable required is %.2lf.\n",min);
    }
    return 0;
}